Question

At $ 25^\circ $, the dissociation constant of a base, BOH is $ 1.0 \times 10^{-12} $. The concentration of hydroxyl ions in 0.01 M aqueous solution of the base would be

• A. $ 2.0 \times 10^{-6} mol L^{-1} $
• B. $ 1.0 \times 10^{-5} mol L^{-1} $
• C. $ 1.0 \times 10^{-6} mol L^{-1} $
• D. $ 1.0 \times 10^{-7} mol L^{-1} $

Answer 1

The Correct Option is D

To find the concentration of hydroxyl ions in the 0.01 M aqueous solution of the base BOH, we need to apply the concept of dissociation constant for a base.

The base dissociation constant K_b is given for BOH, and its value at 25^\circ \text{C} is 1.0 \times 10^{-12}.

BOH dissociates in water as follows:

\text{BOH} \rightleftharpoons \text{B}^+ + \text{OH}^−

For this dissociation, the base dissociation constant K_b is given by:

K_b = \frac{[\text{B}^+][\text{OH}^-]}{[\text{BOH}]}

Let's assume x is the concentration of \text{OH}^− ions at equilibrium, and since BOH dissociates minimally, the equilibrium concentration of BOH can be approximated as the initial concentration minus x:

Initial concentration of BOH = 0.01 \, \text{M}

The equation for the dissociation constant can be written as:

K_b = \frac{x^2}{0.01 - x} \approx \frac{x^2}{0.01}

Since x is small, 0.01 - x \approx 0.01.

Substitute K_b = 1.0 \times 10^{-12} into the equation:

1.0 \times 10^{-12} = \frac{x^2}{0.01}

Now solve for x:

x^2 = (1.0 \times 10^{-12}) \times 0.01

x^2 = 1.0 \times 10^{-14}

Taking the square root of both sides, we get:

x = \sqrt{1.0 \times 10^{-14}} = 1.0 \times 10^{-7} \, \text{mol L}^{-1}

Therefore, the concentration of hydroxyl ions in the 0.01 M aqueous solution of the base BOH is 1.0 \times 10^{-7} \, \text{mol L}^{-1}, which matches the correct answer:

1.0 \times 10^{-7} \, \text{mol L}^{-1}